Question: Simplify the following expression: $y = \dfrac{3x^2+11x- 4}{x + 4}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(3)}{(-4)} &=& -12 \\ {a} + {b} &=& &=& {11} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-12$ and add them together. Remember, since $-12$ is negative, one of the factors must be negative. The factors that add up to ${11}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${12}$ $ \begin{eqnarray} {ab} &=& ({-1})({12}) &=& -12 \\ {a} + {b} &=& {-1} + {12} &=& 11 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({3}x^2 {-1}x) + ({12}x {-4}) $ Factor out the common factors: $ x(3x - 1) + 4(3x - 1)$ Now factor out $(3x - 1)$ $ (3x - 1)(x + 4)$ The original expression can therefore be written: $ \dfrac{(3x - 1)(x + 4)}{x + 4}$ We are dividing by $x + 4$ , so $x + 4 \neq 0$ Therefore, $x \neq -4$ This leaves us with $3x - 1; x \neq -4$.